Calculus Examples

Find the Absolute Max and Min over the Interval f(x)=(x^2-9)^2 on -3 , 3
f(x)=(x2-9)2f(x)=(x29)2 on -33 , 33
Step 1
Find the critical points.
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Step 1.1
Find the first derivative.
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Step 1.1.1
Find the first derivative.
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Step 1.1.1.1
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f(g(x))g(x) where f(x)=x2 and g(x)=x2-9.
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Step 1.1.1.1.1
To apply the Chain Rule, set u as x2-9.
ddu[u2]ddx[x2-9]
Step 1.1.1.1.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=2.
2uddx[x2-9]
Step 1.1.1.1.3
Replace all occurrences of u with x2-9.
2(x2-9)ddx[x2-9]
2(x2-9)ddx[x2-9]
Step 1.1.1.2
Differentiate.
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Step 1.1.1.2.1
By the Sum Rule, the derivative of x2-9 with respect to x is ddx[x2]+ddx[-9].
2(x2-9)(ddx[x2]+ddx[-9])
Step 1.1.1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
2(x2-9)(2x+ddx[-9])
Step 1.1.1.2.3
Since -9 is constant with respect to x, the derivative of -9 with respect to x is 0.
2(x2-9)(2x+0)
Step 1.1.1.2.4
Simplify the expression.
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Step 1.1.1.2.4.1
Add 2x and 0.
2(x2-9)(2x)
Step 1.1.1.2.4.2
Multiply 2 by 2.
4(x2-9)x
4(x2-9)x
4(x2-9)x
Step 1.1.1.3
Simplify.
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Step 1.1.1.3.1
Apply the distributive property.
(4x2+4-9)x
Step 1.1.1.3.2
Apply the distributive property.
4x2x+4-9x
Step 1.1.1.3.3
Combine terms.
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Step 1.1.1.3.3.1
Raise x to the power of 1.
4(x1x2)+4-9x
Step 1.1.1.3.3.2
Use the power rule aman=am+n to combine exponents.
4x1+2+4-9x
Step 1.1.1.3.3.3
Add 1 and 2.
4x3+4-9x
Step 1.1.1.3.3.4
Multiply 4 by -9.
f(x)=4x3-36x
f(x)=4x3-36x
f(x)=4x3-36x
f(x)=4x3-36x
Step 1.1.2
The first derivative of f(x) with respect to x is 4x3-36x.
4x3-36x
4x3-36x
Step 1.2
Set the first derivative equal to 0 then solve the equation 4x3-36x=0.
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Step 1.2.1
Set the first derivative equal to 0.
4x3-36x=0
Step 1.2.2
Factor the left side of the equation.
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Step 1.2.2.1
Factor 4x out of 4x3-36x.
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Step 1.2.2.1.1
Factor 4x out of 4x3.
4x(x2)-36x=0
Step 1.2.2.1.2
Factor 4x out of -36x.
4x(x2)+4x(-9)=0
Step 1.2.2.1.3
Factor 4x out of 4x(x2)+4x(-9).
4x(x2-9)=0
4x(x2-9)=0
Step 1.2.2.2
Rewrite 9 as 32.
4x(x2-32)=0
Step 1.2.2.3
Factor.
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Step 1.2.2.3.1
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=x and b=3.
4x((x+3)(x-3))=0
Step 1.2.2.3.2
Remove unnecessary parentheses.
4x(x+3)(x-3)=0
4x(x+3)(x-3)=0
4x(x+3)(x-3)=0
Step 1.2.3
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
x=0
x+3=0
x-3=0
Step 1.2.4
Set x equal to 0.
x=0
Step 1.2.5
Set x+3 equal to 0 and solve for x.
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Step 1.2.5.1
Set x+3 equal to 0.
x+3=0
Step 1.2.5.2
Subtract 3 from both sides of the equation.
x=-3
x=-3
Step 1.2.6
Set x-3 equal to 0 and solve for x.
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Step 1.2.6.1
Set x-3 equal to 0.
x-3=0
Step 1.2.6.2
Add 3 to both sides of the equation.
x=3
x=3
Step 1.2.7
The final solution is all the values that make 4x(x+3)(x-3)=0 true.
x=0,-3,3
x=0,-3,3
Step 1.3
Find the values where the derivative is undefined.
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Step 1.3.1
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Step 1.4
Evaluate (x2-9)2 at each x value where the derivative is 0 or undefined.
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Step 1.4.1
Evaluate at x=0.
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Step 1.4.1.1
Substitute 0 for x.
((0)2-9)2
Step 1.4.1.2
Simplify.
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Step 1.4.1.2.1
Raising 0 to any positive power yields 0.
(0-9)2
Step 1.4.1.2.2
Subtract 9 from 0.
(-9)2
Step 1.4.1.2.3
Raise -9 to the power of 2.
81
81
81
Step 1.4.2
Evaluate at x=-3.
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Step 1.4.2.1
Substitute -3 for x.
((-3)2-9)2
Step 1.4.2.2
Simplify.
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Step 1.4.2.2.1
Raise -3 to the power of 2.
(9-9)2
Step 1.4.2.2.2
Subtract 9 from 9.
02
Step 1.4.2.2.3
Raising 0 to any positive power yields 0.
0
0
0
Step 1.4.3
Evaluate at x=3.
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Step 1.4.3.1
Substitute 3 for x.
((3)2-9)2
Step 1.4.3.2
Simplify.
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Step 1.4.3.2.1
Raise 3 to the power of 2.
(9-9)2
Step 1.4.3.2.2
Subtract 9 from 9.
02
Step 1.4.3.2.3
Raising 0 to any positive power yields 0.
0
0
0
Step 1.4.4
List all of the points.
(0,81),(-3,0),(3,0)
(0,81),(-3,0),(3,0)
(0,81),(-3,0),(3,0)
Step 2
Evaluate at the included endpoints.
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Step 2.1
Evaluate at x=-3.
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Step 2.1.1
Substitute -3 for x.
((-3)2-9)2
Step 2.1.2
Simplify.
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Step 2.1.2.1
Raise -3 to the power of 2.
(9-9)2
Step 2.1.2.2
Subtract 9 from 9.
02
Step 2.1.2.3
Raising 0 to any positive power yields 0.
0
0
0
Step 2.2
Evaluate at x=3.
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Step 2.2.1
Substitute 3 for x.
((3)2-9)2
Step 2.2.2
Simplify.
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Step 2.2.2.1
Raise 3 to the power of 2.
(9-9)2
Step 2.2.2.2
Subtract 9 from 9.
02
Step 2.2.2.3
Raising 0 to any positive power yields 0.
0
0
0
Step 2.3
List all of the points.
(-3,0),(3,0)
(-3,0),(3,0)
Step 3
Compare the f(x) values found for each value of x in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest f(x) value and the minimum will occur at the lowest f(x) value.
Absolute Maximum: (0,81)
Absolute Minimum: (-3,0),(3,0)
Step 4
 [x2  12  π  xdx ]